Suppose we have a normal projective surface $X$ over an algebraically closed field with 'nice' singularities (say canonical, or perhaps rational Gorenstein, or some other condition), with minimal resolution $Y \rightarrow X$, can we determine the arithmetic genus of a curve $C \subset X$ from numerical information about its strict transform on $Y$, for example the arithmetic genus and the intersection with the exceptional subset?

I know that by adjunction the dualizing sheaf of $C$ is given by $\omega_C =\mathcal{Ext}^1(\mathcal{O}_C, \omega_X$) and that for rational singularities we have that $\omega_X$ is the pushforward of $\omega_Y$, so it seems that perhaps it is possible to find $H^0$ of this sheaf from information on $Y$, but I'm having trouble untwisting the definitions...

If $C$ is a Cartier divisor, then this can be computed from the arithmetic genus of its pull-back. Rational surface singularities are $\mathbb Q$-factorial, so some multiple of $C$ is Cartier, so we can determine the genus of that. I am not sure how to go from Cartier to $\mathbb Q$-Cartier without knowing something about the sheaves $\mathscr{I^m/I^{m+1}}$ where $\mathscr I$ is the ideal sheaf of $C$. If you're interested, I can write up the Cartier case.
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Sándor KovácsDec 14 '11 at 21:21

Thanks Sándor. I think I understand the Cartier case from your discussion on the thread: mathoverflow.net/questions/72353/…. I'm interested if there is anything that can be said for Weil divisors, but it appears some more knowledge of the ideal sheaf is necessary?
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JordanDec 15 '11 at 5:12

It is relatively easy to handle the case when $\mathscr{I/I^2}$ is locally free, but that only happens when $C$ is Cartier.If you knew something about the deformation theory of multiples of $C$ inside $X$ that might help. If there was a reasonable Riemann-Roch, one could connect the arithmetic genus of a multiple to that of the curve. The MO question you are referring to is exactly about a generalized RR, but the problem is that it is still only for line bundles. Any idea I can think of circles back to the Cartier case. Then again, it might be just me and there is solution I'm not thinking of.
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Sándor KovácsDec 15 '11 at 5:36

2 Answers
2

First let us not assume that $X$ has rational singularities, just that it is a normal projective surface and $C\subset X$ a curve on $X$. Let $f:Y\to X$ be a resolution and
$\widetilde C=f^{-1}_*C\subset Y$ the strict transform of $C$ on $Y$.

Since $X$ is normal, $\beta$ is an isomorphism and it is clear that $\alpha$ and $\gamma$ are injective. Then by the Snake Lemma $\alpha$ is also an isomorphism and we obtain that we have an exact sequence
$$
0\to \mathscr O_C \to f_*\mathscr O_C \to R^1f_*\mathscr I_{\widetilde C}\to R^1f_*\mathscr O_{Y}\to 0
$$

Obviously the two $R^1$ sheaves are supported on a zero-dimensional scheme $P$, so we have
$$
\chi(\mathscr O_C)=\chi(\mathscr O_{\widetilde C})-\mathrm{length}(R^1f_*\mathscr I_{\widetilde C}) + \mathrm{length}(R^1f_*\mathscr O_{Y})
$$

Now

1 If $(X,C)$ is log canonical, then it is a DB pair by Thm 1.4 of this paper and Prop 5.1 of this paper. Then $R^1f_*\mathscr I_{\widetilde C}=0$ by Cor 6.2 of the same paper. Therefore in this case $\chi(\mathscr O_C)=\chi(\mathscr O_{\widetilde C})$. But of course, I just realize now that this simply means that if $(X,C)$ is lc, then at every point either $X$ is smooth or $C$ is smooth, so this is not a big surprise.

2 If $X$ has rational singularities, then $R^1f_*\mathscr O_{Y}=0$, so the question reduces to determining $R^1f_*\mathscr I_{\widetilde C}$. In this case this is the same as Sasha's sheaf $T$. Clearly, it feels that one should be able to compute this by knowing the intersection of the exceptional set with $\widetilde C$.

In these situations one may try to use the Theorem on Formal Functions. That says that if $E$ denotes the pre-image of the reduced scheme supported on the singular set of $X$ (i.e., for each singular point the pre-image of the closed point defined by the maximal ideal, sometimes this is called the Artin cycle), then
$$
(R^1f_*\mathscr I_{\widetilde C})_P^{\wedge}\simeq \underset{\leftarrow}{\lim}\ H^1(mE,\mathscr I_{\widetilde C}\otimes \mathscr O_{mE})
$$

Fortunately the rest of the computation is happening on $Y$ so every divisor is Cartier.
I would try to compute the right hand side using the short exact sequences

where $\mathscr L=\mathscr I_{E}/\mathscr I_{E}^2=\mathscr N_{E/Y}^{-1}$ the dual of the normal bundle of $E$ which is indeed a line bundle since $E$ is a local complete intersection and for the same reason $\mathscr I_{E}^m/\mathscr I_{E}^{m+1}\simeq \mathscr L^m$.

Again, $\mathscr I_{\widetilde C}$ is a line bundle on $Y$ and remains that when restricted to $E$, so $(\star)$ induces the short exact sequence

$\mathscr L$ is ample on $E$, so by Serre vanishing $H^1(E, \mathscr I_{\widetilde C}\otimes\mathscr L^{m})=0$ for $m\gg 0$, so this is indeed a finite game.

Another thing that could help is that
by Thms 3 and 4 of this paper $h^0(E,\omega_E)=0$ and hence
$h^1(E, \mathscr I_{\widetilde C}\otimes\mathscr L^{m}) =h^0(E,\omega_E\otimes \mathscr I_{\widetilde C}^{-1}\otimes\mathscr L^{-m})=0$ as soon as $\mathscr L^m(-\widetilde C)$ has a global section. You should be able to figure out the first $m$ for which this happens as that depends on $E^2$ (see Thm 4 of ibid) and $\widetilde C\cdot E$. For the actual dimension of the intermediate steps you might be able to use some of Artin's methods from here.

This has certainly grown to a much longer answer than I anticipated when I started and although it does not give you a complete answer it might give you some ideas to go on.

Let $C'$ be the strict transform. Then there is an exact sequence
$$
0 \to O_C \to f_*O_{C'} \to T \to 0,
$$
where $T$ is a sheaf supported at the image of the exceptional set.
This siquence gives $\chi(O_C) = \chi(O_{C'}) - \ell(T)$, so you only have to know the length of $T$.

In some sense $T$ ``measures'' the intersection of $C'$ with the intersection set, but you should be accurate here. For example, if $C'$ intersects a component of the exceptional set transversally at $k$ points, then $T$ is of length $k-1$ at the image of this component. However, if $C'$ is singular at some point of intersection with the exceptional set, the effect may be more complicated.